Type丶Alpha 2016-01-07 09:24 采纳率: 100%
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关于那个长整数的加法的代码,能不能帮忙注释一下每句的作用

#include
#include
#include
typedef struct node{
int e;
struct node *next;
struct node *prir;
} node;
main()
{
node *num1,*num2,*sum;
node *create(node *n);
int output(node *n);
node *add_Num(node *n1,node *n2);
num1=(node *)malloc(sizeof(node));
num2=(node *)malloc(sizeof(node));
num1=create(num1);
num2=create(num2);
sum=add_Num(num1,num2);
output(sum);
}
int output(node *n)
{
node *p;
if(n->e printf("-");
printf("%d,",n->next->e);
p=n->next->next;
while(p!=n->next)
{
if(p->e if(p->e if(p->e printf("%d,",p->e);
p=p->next;
}
printf("\b \b\n");
}

node *create(node *n)
{
node *p,*t;
int e,i;
t=(node *)malloc(sizeof(node));
i=scanf("%d",&e);
if(ee=-1;
else n->e=1;
t->e=n->e*e;
n->next=t;
t->next=t;
t->prir=t;
n->prir=NULL;
p=n->next;
while(i!=0)
{
if(getchar()==',')
i=scanf("%d",&e);
else break;
t=(node *)malloc(sizeof(node));
t->e=e;
p->next=t;
t->prir=p;
t->next=n->next;
n->next->prir=t;
p=p->next;
}
return(n);
}
node *add_Num(node *n1,node *n2)
{
node *p1,*p2,*t,*s,*sum;
int e1,e2,carry;
p1=n1->next->prir;
p2=n2->next->prir;
e1=n1->e;
e2=n2->e;
sum=(node *)malloc(sizeof(node));
sum->e=1;
sum->prir=NULL;
s=(node *)malloc(sizeof(node));
s->e=p1->e*e1+p2->e*e2;
p1=p1->prir;
p2=p2->prir;
if(e1*e2==1)
{
sum->e=e1;
s->e=abs(s->e);
if(s->e>=10000)
{
s->e-=10000;
carry=1;
}
else carry=0;
}
else
if(s->e>=0) carry=0;
else
{
s->e+=10000;
carry=-1;
}
t=s;
s->next=s;
s->prir=s;
sum->next=s;
while(p1!=n1->next->prir&&p2!=n2->next->prir)
{
s=(node *)malloc(sizeof(node));
s->e=p1->e*e1+p2->e*e2+sum->e*carry;
p1=p1->prir;
p2=p2->prir;
if(e1*e2==1)
{
sum->e=e1;
s->e=abs(s->e);
if(s->e>=10000)
{
s->e-=10000;
carry=1;
}
else carry=0;
}
else
if(s->e>=0) carry=0;
else
{
s->e+=10000;
carry=-1;
}
s->next=t;
t->prir=s;
s->prir=sum->next;
sum->next->next=s;
t=t->prir;
}

if(p1==n1->next->prir)
while(p2!=n2->next->prir)
{
s=(node *)malloc(sizeof(node));
s->e=p2->e*e2+sum->e*carry;
p2=p2->prir;
if(e1*e2==1)
{
sum->e=e1;
s->e=abs(s->e);
if(s->e>=10000)
{
s->e-=10000;
carry=1;
}
else carry=0;
}
else
if(s->e>=0) carry=0;
else
{
s->e+=10000;
carry=-1;
}
s->next=t;
t->prir=s;
s->prir=sum->next;
sum->next->next=s;
t=t->prir;
}
if(p2==n2->next->prir)
while(p1!=n1->next->prir)
{
s=(node *)malloc(sizeof(node));
s->e=p1->e*e1+sum->e*carry;
p1=p1->prir;
if(e1*e2==1)
{
sum->e=e1;
s->e=abs(s->e);
if(s->e>=10000)
{
s->e-=10000;
carry=1;
}
else carry=0;
}
else
if(s->e>=0) carry=0;
else
{
s->e+=10000;
carry=-1;
}
s->next=t;
t->prir=s;
s->prir=sum->next;
sum->next->next=s;
t=t->prir;
}
if(carry!=0)
{
s=(node *)malloc(sizeof(node));
s->e=1;
carry=0;
s->next=t;
t->prir=s;
s->prir=sum->next;
sum->next->next=s;
t=t->prir;
}
sum->next=t;
do
{
if(t->e==0)
{
sum->next=t->next;
t->next->prir=t->prir;
t->prir->next=t->next;
free(t);
}
else break;
t=sum->next;
}while(t!=sum->next->prir);
return(sum);
}

  • 写回答

3条回答 默认 最新

  • ysuwood 2016-01-07 09:53
    关注

    这么长的东西,让人每一句都注释一下,有点强人所难。建议你先自己研究一下,哪里不懂提出来。

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
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